1. ## Binomial Expansion

Expand the first 3 terms in $\displaystyle [6-13x+6x^2]^5$

2. Originally Posted by Punch
Expand the first 3 terms in $\displaystyle [6-13x+6x^2]^5$
when using expansion of $\displaystyle (a+b)^n$

$\displaystyle T_{k+1}=\displaystyle \binom {n}{k} a^{n-k} \cdot b^{k}$

that's to find any member (first, second, ... )

so u try and see where do u go with this

3. What you can do is treat it like this:

$\displaystyle [6 - (13x - 6x^2)]^5$

Then apply the formula to find the first three terms.

$\displaystyle (a + b)^n = a^n + \binom{n}{1} (a)^{n-1}(b) + \binom{n}{2} (a)^{n-2}(b)^2+...$

Taking a = 6
b = - (13x - 6x^2)

4. that makes sense thanks

5. Originally Posted by Unknown008
What you can do is treat it like this:

$\displaystyle [6 - (13x - 6x^2)]^5$

Then apply the formula to find the first three terms.

$\displaystyle (a + b)^n = a^n + \binom{n}{1} (a)^{n-1}(b) + \binom{n}{2} (a)^{n-2}(b)^2+...$

Taking a = 6
b = - (13x - 6x^2)
perhaps that's the better way to go

formula that i posted is the same, but way that it's wrote it's way easier to find let's say 181 element of some expansion not for this problem but works the same

6. Originally Posted by Punch
Expand the first 3 terms in $\displaystyle [6-13x+6x^2]^5$
I have a question about the question itself.
Who is to say which are the first three terms?

7. Originally Posted by Plato
I have a question about the question itself.
Who is to say which are the first three terms?
Yes, the question should have included something like: "the first three terms in ascending powers of x"

8. Originally Posted by Plato
I have a question about the question itself.
Who is to say which are the first three terms?
I was wondering that too, figured the only natural choices were: (1) constant term, x^1 term, x^2 term; (2) x^10 term, x^9 term, x^8 term; and that since the trinomial we are expanding is given in form (1) it would be acceptable to assume form (1).

9. Originally Posted by Punch
Expand the first 3 terms in $\displaystyle [6-13x+6x^2]^5$
You can also expand

$\displaystyle \left(x[6x-13]+6\right)^5$

It may be more compact.

10. Hello, Punch!

This is a messy one . . . even with a good procedure.

$\displaystyle \text{Find the first 3 terms of: }\;(6-13x+6x^2)^5$

It is evident that the first three terms are:
. . . . $\displaystyle \text{constant},\;x\text{-term},\;x^2\text{-term}$

$\displaystyle \text{We have: }\:(a + b + c)^5\quad \text{where: }\:\begin{Bmatrix}a &=& 6 \\ b &=& \text{-}13x \\ c &=& 6x^2 \end{Bmatrix}$

The "Trinomial Expansion" forms an array:

$\displaystyle \begin{array}{cccccccccccc}&&& {5\choose5,0,0}a^5 \\ && {5\choose4,1,0}a^4b && {5\choose4,0,1}a^4c \\ & {5\choose3,2,0}a^3b^2 && {5\choose3,1,1}a^3bc && {5\choose3,0,2}a^3c^2 \\ {5\choose2,3,0}a^2b^3 && {5\choose2,2,1}a^2b^2c && {5\choose2,1,2}a^2bc^2 && {5\choose2,0,3}a^2c^3 \\ \vdots && \vdots && \vdots && \vdots\end{array}$

We want only the first four terms (reading from the top).
. . [The rest contain $\displaystyle x^3$ or higher powers.]

. . $\displaystyle {5\choose5,0,0}(6^5) + {5\choose4,1,0}(6^4)(\text{-}13x) + {5\choose4,0,1}(6^4)(6x^2) + {5\choose3,2,0}(6^3)(\text{-}13x)^2$

. . $\displaystyle =\;1(7776) + 5(1296)(\text{-}13x) + 5(1296)(6x^2) + 10(216)(169x^2)$

. . $\displaystyle =\;7,\!776 - 84,\!240x + 38,\!880x^2 + 365,\!040x^2$

. . $\displaystyle =\;7,\!776 - 84,\!240x + 403,\!920x^2$

But check my work . . . please!

11. or look at it like this

$\displaystyle (6-13x+6x^2)^5=[-x+(x-1)(6x-6)]^5$

and so when doing u'll get

$\displaystyle T_1= \displastyle \binom {5}{0} (-x)^5 [(x-1)(6x-6)]^0=-x^5$

$\displaystyle T_2= \displastyle \binom {5}{1} (-x)^4 [(x-1)(6x-6)]^1=5x^4[6x^2-12x+6]=30x^6-60x^5+30x^4$

$\displaystyle T_3= \displastyle \binom {5}{2} (-x)^3 [(x-1)(6x-6)]^2=-10x^3[(x-1)^2(6x-6)^2]=-360x^7+1440x^6-2160x^5+1440x^4-360x^3$

u can go further... but u get the idea

or am i wrong here ?

12. Originally Posted by Soroban
. . $\displaystyle =\;7,\!776 - 84,\!240x + 403,\!920x^2$

But check my work . . . please!
Confirmed with Mathematica. Unknown008's method gives the same result.

13. Originally Posted by yeKciM
or look at it like this

$\displaystyle (6-13x+6x^2)^5=[-x+(x-1)(6x-6)]^5$

and so when doing u'll get

$\displaystyle T_1= \displastyle \binom {5}{0} (-x)^5 [(x-1)(6x-6)]^0=-x^5$

$\displaystyle T_2= \displastyle \binom {5}{1} (-x)^4 [(x-1)(6x-6)]^1=5x^4[6x^2-12x+6]=30x^6-60x^5+30x^4$

$\displaystyle T_3= \displastyle \binom {5}{2} (-x)^3 [(x-1)(6x-6)]^2=-10x^3[(x-1)^2(6x-6)^2]$
$\displaystyle =-360x^7+1440x^6-2160x^5+1440x^4-360x^3$

u can go further... but u get the idea

or am i wrong here ?
This makes it harder to find the first three terms, compared with what Unknown008 posted, doesn't it?

14. Originally Posted by undefined
This makes it harder to find the first three terms, compared with what Unknown008 posted, doesn't it?

hmmmm

probably lost in translation, but if u look at $\displaystyle (6-13x+6x^2)^5=[-x+(x-1)(6x-6)]^5$ then first one should be :

$\displaystyle T_1=\displaystyle \binom{5}{0} a^{5-0}\cdot b^0$

and so on...

but probably i'm little (or a lot) of with meaning of "terms" for me it's like n-th member or something like that

Edit: Sorry, but now i'm little a bit of confused

15. Originally Posted by yeKciM
hmmmm

probably lost in translation, but if u look at $\displaystyle (6-13x+6x^2)^5=[-x+(x-1)(6x-6)]^5$ then first one should be :

$\displaystyle T_1=\displaystyle \binom{5}{0} a^{5-0}\cdot b^0$

and so on...

but probably i'm little (or a lot) of with meaning of "terms" for me it's like n-th member or something like that

Edit: Sorry, but now i'm little a bit of confused
There are 11 terms in the expansion of $\displaystyle (6-13x+6x^2)^5$, starting with the constant term and ending with the $\displaystyle x^{10}$ term. Soroban gave the first three terms in post #10.

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